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题目链接： http://www.lightoj.com/volume_showproblem.php?problem=1005

代码：

//先在n行中选出k行，C(n,k)，再在n列中选出k列随便放A(n,k)，答案为C(n,k)*A(n,k)
#include<stdio.h>     
#include<iostream>     
#include<math.h>     
#include<stdlib.h>     
#include<ctype.h>     
#include<algorithm>     
#include<vector>     
#include<string.h>     
#include<stack>     
#include<set>     
#include<map>     
#include<sstream>     
#include<time.h>     
#include<utility>     
#include<malloc.h>     
#include<stdexcept>     
#include<iomanip>     
#include<iterator>     

using namespace std;

int t;
int n, k;

long long a[35][35], c[35][35];

void solve()
{
    for (int i = 1; i <= 30; i++)
    {
        c[i][0] = c[i][i] = 1;
        for (int j = 1; j < i; j++)
        {
            c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
        }
    }
    for (int i = 0; i <= 30; i++)
    {
        a[i][0] = 1;
        a[i][1] = i;
        for (int j = 2; j <= i; j++)
            a[i][j] = a[i][j - 1] * (long long)(i - j + 1);
    }
}

int main()
{
    int cases = 1; 
    scanf("%d",&t);
    solve();
    while (t--)
    {
        scanf("%d%d", &n, &k);
        printf("Case %d: ", cases++);
        long long ans = 0 ;
        if (k > n)  { printf("0\n"); continue; }
        ans = a[n][k] * c[n][k];
        printf("%lld\n",ans);
    }
    return 0;
}